$\int e^{2x} (\sin x +\frac{1}{2} \cos x) dx$ is equal to

$\frac{1}{2}e^{2x}\sin x+C$, C is an arbitrary constant

The correct answer is Option (2) → $\frac{1}{2}e^{2x}\sin x+C$, C is an arbitrary constant

$I = \int e^{2x}\left(\sin x + \frac{1}{2}\cos x\right)dx$

$I = \int e^{2x}\sin x\,dx + \frac{1}{2}\int e^{2x}\cos x\,dx$

Using standard results:

$\int e^{ax}\sin bx\,dx = \frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx - b\cos bx)$

$\int e^{ax}\cos bx\,dx = \frac{e^{ax}}{a^{2}+b^{2}}(a\cos bx + b\sin bx)$

Substitute $a=2,\;b=1$:

$I = \frac{e^{2x}}{5}(2\sin x - \cos x) + \frac{1}{2}\cdot\frac{e^{2x}}{5}(2\cos x + \sin x) + C$

$I = \frac{e^{2x}}{10}\left[(4\sin x - 2\cos x) + (2\cos x + \sin x)\right] + C$

$I = \frac{e^{2x}}{10}(5\sin x) + C$

$I = \frac{1}{2}e^{2x}\sin x + C$